Continued fraction to irrational number Let $[1;\overline{2,1}$] be a continued fraction. I want to find the corresponding number. I know how to transform fractions of the form $[a;\overline{b}]$ but I am having a hard time here. Thank you.
 A: The idea is not much different. For something like this, it becomes $$ \left[ 1; \overline {2, 1} \right] = 1 + \cfrac {1}{2 + \cfrac {1}{1 + \cfrac {1}{2 + \cfrac {1}{1 + \cdots}}}} $$Then, you can use the regular technique. If this is $x$, then $$ x - 1 = \cfrac {1}{2 + \cfrac {1}{x}} \implies x - 1 = \cfrac {1}{\cfrac {2x + 1}{x}} = \frac {x}{2x+1}, $$then you can solve. 
In general, for $\overline{b_1,b_2,\cdots,b_n}$, instead of the denominators outside of the infinite fractions being continues $b$ as suggested by $\overline{b}$, it keeps cycling between $b_1,b_2,b_3,\cdots,b_n$. 
A: Let $$A = 1+ \cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{\ddots}}}}$$
Then notice that $$A=1+\cfrac{1}{2+\cfrac{1}{A}}$$
This gives
$$(A-1)(2A+1)=A$$
You can solve this quadratic :)

Notice that this method works for many types of infinite and repeating expressions. For example, this nested square root thing:
$$B=3+\sqrt{3+\sqrt{3+\dots}} \qquad \Rightarrow\qquad (B-3)^2=B$$
The geometric series is another example ($|a|<1$ of course):
$$B=a+a^2+a^3+\dots\qquad\Rightarrow\qquad B-a=aB$$
